# Integral Test

This page has been identified as a candidate for refactoring of medium complexity.In particular: Renamed from Euler-Maclaurin Summation Formula. Work in progress setting that one up.Until this has been finished, please leave
`{{Refactor}}` in the code.
Because of the underlying complexity of the work needed, it is recommended that you do not embark on a refactoring task until you have become familiar with the structural nature of pages of $\mathsf{Pr} \infty \mathsf{fWiki}$.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Refactor}}` from the code. |

## Theorem

Let $f$ be a real function which is continuous, positive and decreasing on the interval $\hointr 1 {+\infty}$.

Let the sequence $\sequence {\Delta_n}$ be defined as:

- $\ds \Delta_n = \sum_{k \mathop = 1}^n \map f k - \int_1^n \map f x \rd x$

Then $\sequence {\Delta_n} $ is decreasing and bounded below by zero.

Hence it converges.

## Proof

From Upper and Lower Bounds of Integral, we have that:

- $\ds m \paren {b - a} \le \int_a^b \map f x \rd x \le M \paren {b - a}$

where:

- $M$ is the maximum

and:

- $m$ is the minimum

of $\map f x$ on $\closedint a b$.

Since $f$ decreases, $M = \map f a$ and $m = \map f b$.

Thus it follows that:

- $\ds \forall k \in \N_{>0}: \map f {k + 1} \le \int_k^{k + 1} \map f x \rd x \le \map f k$

as $\paren {k + 1} - k = 1$.

Thus:

\(\ds \Delta_{n + 1} - \Delta_n\) | \(=\) | \(\ds \paren {\sum_{k \mathop = 1}^{n + 1} \map f k - \int_1^{n + 1} \map f x \rd x} - \paren {\sum_{k \mathop = 1}^n \map f k - \int_1^n \map f x \rd x}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \map f {n + 1} - \int_n^{n + 1} \map f x \rd x\) | ||||||||||||

\(\ds \) | \(\le\) | \(\ds \map f {n + 1} - \map f {n + 1}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 0\) |

Thus $\sequence {\Delta_n}$ is decreasing.

$\Box$

Also:

\(\ds \Delta_n\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \map f k - \sum_{k \mathop = 1}^{n - 1} \int_k^{k + 1} \map f x \rd x\) | ||||||||||||

\(\ds \) | \(\ge\) | \(\ds \sum_{k \mathop = 1}^n \map f k - \sum_{k \mathop = 1}^{n - 1} \map f k\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \map f n\) | ||||||||||||

\(\ds \) | \(\ge\) | \(\ds 0\) |

$\Box$

Hence the result.

$\blacksquare$

## Also known as

The **integral test** is also known as the **Euler-Maclaurin Summation Formula**, but that result properly refer to a more precise theorem of which this is a simple corollary.

## Notes

It follows from this that if $f$ is continuous, positive and decreasing on $\hointr 1 \infty$, then the series $\ds \sum_{k \mathop = 1}^\infty \map f k$ and the improper integral $\ds \int_1^{\mathop \to +\infty} \map f x \rd x$ either both converge or both diverge.

So this theorem provides a test for the convergence of both a series and an improper integral.

This page or section has statements made on it that ought to be extracted and proved in a Theorem page.In particular: The above statement deserves to be put into a page of its own, or included as a corollary, rather than relegated to the notes.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by creating any appropriate Theorem pages that may be needed.To discuss this page in more detail, feel free to use the talk page. |

## Sources

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 13.32$ - 1992: Larry C. Andrews:
*Special Functions of Mathematics for Engineers*(2nd ed.) ... (previous) ... (next): $\S 1.2.2$: Summary of convergence tests: Theorem $1.2$